Given an array of n integers where n > 1, nums, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].

Solve it without division and in O(n).

For example, given [1,2,3,4], return [24,12,8,6].
方法一、

vector<int> productExceptSelf(vector
 
<int>& nums) {
        int len = nums.size();
        vector<int> left,right(len),ret(len);

        left.push_back(1);
        for(int i=1; i<len; i++){
            left.push_back(left[i-1]*nums[i-1]);
        }

        right[len-1]=1;
        for(int i=len-2; i>=0; i--){
            right[i]=right[i+1
 
]*nums[i+1];
        }

        for(int i=0; i<len; i++){
            ret[i]=left[i]*right[i];
        }

        return ret;
    }

方法二、O(n) time and O(n) space solution.

vector<int> productExceptSelf(vector<int>& nums) {
        int n=nums.size();
        vector<int> fromBegin(n);
        fromBegin[0
 
]=1;
        vector<int> fromLast(n);
        fromLast[0]=1;

        for(int i=1;i<n;i++){
            fromBegin[i]=fromBegin[i-1]*nums[i-1];
            fromLast[i]=fromLast[i-1]*nums[n-i];
        }

        vector<int> res(n);
        for(int i=0;i<n;i++){
            res[i]=fromBegin[i]*fromLast[n-1-i];
        }
        return res;
    }

方法三、

vector<int> productExceptSelf(vector<int>& nums) {
        int n=nums.size();
        int fromBegin=1;
        int fromLast=1;
        vector<int> res(n,1);

        for(int i=0;i<n;i++){
            res[i]*=fromBegin;
            fromBegin*=nums[i];
            res[n-1-i]*=fromLast;
            fromLast*=nums[n-1-i];
        }
        return res;
    }